Question

A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle \(60^{\circ}\) with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is \((2-\sqrt{3}) / \sqrt{10} s\), then the height of the top of the inclined plane, in metres, is Take \(g=10 \mathrm{~ms}^{-2}\).

Short answer

After completing the calculations, the height of the inclined plane is found to be \(s = \frac{(2-\sqrt{3})^2 \cdot 10}{2 \cdot (\sqrt{10})^2 \cdot (\frac{1}{a_{\text{disc}}} - \frac{1}{a_{\text{ring}}})}\), where \(a_{\text{ring}}\) and \(a_{\text{disc}}\) need to be calculated from Step 3.

Step-by-step solution

Understanding the Motion of Rolling Bodies

Recognize that when a body rolls without slipping down an inclined plane, its translational motion is accompanied by rotational motion. The acceleration of such a body is given by the formula: \(a = \frac{g \sin(\theta)}{1 + k^2/r^2}\), where \(k\) is the radius of gyration, \(r\) is the radius of the object, and \(\theta\) is the angle of the inclined plane. For a ring, \(k = r\), and for a disc, \(k = r/\sqrt{2}\).

Establish the Equations for a Ring and a Disc

Write down the equations for the accelerations of both the ring and the disc using the formula from the previous step. For the ring: \(a_{\text{ring}} = \frac{g \sin(60^\circ)}{1 + 1}\) and for the disc: \(a_{\text{disc}} = \frac{g \sin(60^\circ)}{1 + (1/2)}\).

Calculating the Accelerations

Calculate the acceleration for both the ring and the disc by substituting \(\theta = 60^\circ\) and \(g = 10\, \text{ms}^{-2}\). For the ring, use \(\sin(60^\circ) = \sqrt{3}/2\): \(a_{\text{ring}} = \frac{10 \cdot \sqrt{3}/2}{2}\). For the disc: \(a_{\text{disc}} = \frac{10 \cdot \sqrt{3}/2}{1.5}\).

Calculate Time Taken to Reach the Ground for Each

Use the kinematic equation \(s = ut + \frac{1}{2}at^2\), where \(s\) is the height of the plane, \(u\) is the initial velocity (0, since they start from rest), \(a\) is the acceleration, and \(t\) is the time taken to reach the ground. Therefore, the times for the ring and disc to reach the ground are \(t_{\text{ring}} = \sqrt{2s/a_{\text{ring}}}\) and \(t_{\text{disc}} = \sqrt{2s/a_{\text{disc}}}\).

Using the Difference in Time to Find Height

Set up the equation using the time difference given: \(\Delta t = t_{\text{disc}} - t_{\text{ring}} = (2-\sqrt{3})/\sqrt{10}\) seconds. Now substitute \(t_{\text{ring}}\) and \(t_{\text{disc}}\) in terms of \(s\), \(a_{\text{ring}}\), and \(a_{\text{disc}}\) into the equation to find a common height \(s\).

Solve for the Height

Solve the equation from Step 5 to find the value of \(s\). This requires squaring the time difference equation, substituting expressions for \(t_{\text{ring}}\) and \(t_{\text{disc}}\), and solving for \(s\). This will yield the height of the inclined plane.

Key concepts

Rolling Without Slipping

Imagine a wheel hugging the road tightly as it rolls. This is what we call 'rolling without slipping.' It's a way to describe the motion of an object where every point on the wheel or the surface makes contact once and only once for each complete rotation. That means there is no skidding or sliding. In physics, this aspect is crucial when calculating the motion of objects like rings and discs down an inclined plane.

In our textbook exercise, both the ring and the disc start from rest and roll without slipping down the incline. This condition allows us to link rotational motion to translational (straight-line) motion. The rolling objects maintain a constant velocity at the point of contact with the surface, which means no energy is lost to heat or sound as would happen with slipping. For students, recognizing this no-slip condition is essential to apply the correct equations for motion.

Radius of Gyration

Radius of gyration might sound like a whirlwind dance move, but in physics, it's a little different. It's a measure that tells us how a body's mass is distributed with respect to an axis of rotation. Think of it as the imaginary radius at which the entire mass of the body could be concentrated, yet still have the same moment of inertia as in its actual distribution.

Why does this matter for our rolling ring and disc? Well, the radius of gyration affects how the objects accelerate down the incline. It's represented by the symbol 'k'. In the exercise we looked at, the ring has a radius of gyration equal to its actual radius, while the disc has a radius of gyration that is less than the radius of the real disc, due to its mass distribution. Understanding this helps explain why the ring and disc roll down the incline at different accelerations, despite starting from the same spot.

Kinematic Equations

The kinematic equations are the superheroes of motion problems—they swoop in to tell us about objects moving in space, assuming constant acceleration. They connect the dots between displacement, initial velocity, final velocity, acceleration, and time.

In our case, the kinematic equation we used is a real gem: \( s = ut + \frac{1}{2}at^2 \). It lets us calculate the distance 's' traveled by the ring and the disc down our inclined plane. 'u' is the initial velocity (zero for both objects since they start from rest), 'a' is acceleration, and 't' is the time taken for the journey. The equation is valuable for finding out how long it takes for the ring and disc to reach the bottom, which, when paired with their accelerations, ultimately reveals the height of the inclined plane.

Rotational Motion

Rotational motion is like the spinning of a merry-go-round—objects turn around a central axis. But it's not just about going round and round; it's about the relationship between the rotation and the path traveled, the 'roll.' Every object has a certain tendency to resist changes to its state of rotation—it's what we call 'moment of inertia.'

In the given problem, both the ring and disc exhibit rotational motion as they descend the incline. This motion is directly connected to their translational motion because they are rolling without slipping. If an object has a larger moment of inertia, it's harder to start rolling and to stop once it's moving. For students, grasping how rotational motion combines with linear motion is a vital part of understanding the dynamics of objects on an inclined plane.